hdu 5067(暴力搜索)

Harry And Dig Machine

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 852    Accepted Submission(s): 348

Problem Description

  As
we all know, Harry Porter learns magic at Hogwarts School. However,
learning magical knowledge alone is insufficient to become a great
magician. Sometimes, Harry also has to gain knowledge from other certain
subjects, such as language, mathematics, English, and even algorithm.
  Dumbledore,
the headmaster of Hogwarts, is planning to construct a new teaching
building in his school. The area he selects can be considered as an n*m
grid, some (but no more than ten) cells of which might contain stones.
We should remove the stones there in order to save place for the
teaching building. However, the stones might be useful, so we just move
them to the top-left cell. Taking it into account that Harry learned how
to operate dig machine in Lanxiang School several years ago, Dumbledore
decides to let him do this job and wants it done as quickly as
possible. Harry needs one unit time to move his dig machine from one
cell to the adjacent one. Yet skilled as he is, it takes no time for him
to move stones into or out of the dig machine, which is big enough to
carry infinite stones. Given Harry and his dig machine at the top-left
cell in the beginning, if he wants to optimize his work, what is the
minimal time Harry needs to finish it?

 

Input

They are sever test cases, you should process to the end of file.
For each test case, there are two integers n and m.(1≤n,m≤50).
The next n line, each line contains m integer. The j-th number of ith line a[i][j] means there are a[i][j] stones on the jth cell of the ith line.( 0≤a[i][j]≤100 , and no more than 10 of a[i][j] will be positive integer).

 

Output

For each test case, just output one line that contains an integer indicate the minimal time that Harry can finish his job.

 

Sample Input

3 3
0 0 0
0 100 0
0 0 0
2 2
1 1
1 1

 

Sample Output

4
4

 

题意:n*m的矩阵里面有k个石头(k<=10) 哈利波特从(0,0)点出发,可以沿着x轴或者y轴在矩阵里面行走,他的任务是在最短的时间内将石头全部取完,问至少需要多少时间他才能取完所有的石头然后回到原点？？

题解:构图然后暴搜。。剪枝。。

#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int INF = ;
struct Point
{
    int x,y;
} p[];
int n,m,res,cnt;
int graph[][];
bool vis[];
void dfs(int x,int dept,int ans)
{
    vis[x] = true;
    if(ans>res) return ; ///剪枝
    if(dept==cnt-)
    {
        ans += graph[x][]; ///走回去
        res = min(res,ans);
        return ;
    }
    for(int i=; i<cnt; i++)
    {
        if(!vis[i]&&graph[x][i]!=INF&&x!=i)
        {
            dfs(i,dept+,ans+graph[x][i]);
            vis[i] = false;
        }
    }
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        cnt = ;
        int v;
        for(int i=; i<n; i++)
        {
            for(int j=; j<m; j++)
            {
                scanf("%d",&v);
                if(i==&&j==)
                {
                    p[cnt].x = i,p[cnt++].y=j;
                }
                else if(v)
                {
                    p[cnt].x = i,p[cnt++].y=j;
                }
            }
        }
        for(int i=; i<cnt; i++)
        {
            for(int j=; j<cnt; j++)
            {
                if(i==j)graph[i][j] = ;
                else graph[i][j] =INF;
            }
        }
        for(int i=; i<cnt; i++)
        {
            for(int j=i+; j<cnt; j++)
            {
                graph[i][j] = graph[j][i] = abs(p[i].x-p[j].x)+abs(p[i].y-p[j].y);
            }
        }
        /*for(int i=0; i<cnt; i++)
        {
            for(int j=0; j<cnt; j++)
            {
                printf("%d ",graph[i][j]);
            }
            printf("\n");
        }*/
        memset(vis,false,sizeof(vis));
        res = INF;
        vis[] = true;
        dfs(,,);
        printf("%d\n",res);
    }
    return ;
}